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The vapor pressure of benzaldehyde at 97 °C is about 60 mm Hg. Assume atmospheric pressure is 760 mm Hg. If you collected 20 g of distillate from your steam distillation, how much of it was benzaldehyde? Show your calculation.

The vapor pressure of benzaldehyde at 97 °C is about 60 mm Hg. Assume atmospheric pressure is 760 mm Hg. If you collected 20 g of distillate from your steam distillation, how much of it was benzaldehyde? Show your calculation.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The answer to c is 1.58g. Please answer question on bottom.  

h extraction I never need ne. See the lly enough. his is so sim- a test tube solve in the ottom, float er, dissolve. to be removed d more fresh ave the same e organic is he theory is our organic emembered. he solubility Chapter 13). kind of a fog ended drop- . Or stir the ents of your on-breaking an react with clogs up the anels. he two liquid not enough nt of one layer ore, you may dissolve this rial does not analysis. You 've heard this acid, and you up, you find some solubilities of oxalic acid as follows: 9.5 g/100 g in water; 23.7 g/100 g in etha- nol; 16.9 g/100 g in diethyl ether. Based on the solubilities, you decide to extract into ethanol from water, forgetting for the moment that ethanol is soluble in water and that you must have two insoluble liquids to carry out an extraction. Chagrined, you forget the ethanol and choose diethyl ether. From the preceding solubility data, we can calculate the distribution, or partition coefficient, for oxalic acid in the water-ether extraction. This coefficient (number) is just the ratio of solubilities of the compound you wish to extract in the two lay- ers. Here, K₁ and = solubility of oxalic acid in ether solubility of oxalic acid in water which amounts to 16.9/9.5, or 1.779. Imagine that you have 40 g of oxalic acid in 1000 mL water and you put that in contact with 1000 mL ether. The oxalic acid distributes itself between the two layers. How much is left in each layer? Well, if we let x g equal the amount that stays in the water, 1.779x g of the acid has to walk over to the ether. And so Wt of oxalic acid in ether= (1000 mL)(1.779x g/mL)=1779x g Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g The total weight of the acid is 40 g (now partitioned between two layers) and SO How to Extract and Wash What 2779x g = 40 g x = 0.0144 Wt of oxalic acid in ether = 1779(0.0144) g = 25.6 g Wt of oxalic acid in water = 1000 (0.0144) g = 14.4 g Now, let's start with the same 40 g of oxalic acid in 1000 mL of water, but this time we will do three extractions with 300 mL of ether. The first 300-mL portion hits, and (300 mL) (1.779x g/mL)=533.7x g Wt of oxalic acid in ether = Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g The total weight of the acid is 40 g (now partitioned between two layers) and 1533.7x g = 40 g x = 0.0261 Wt of oxalic acid in ether = 533.7 (0.0261) g = 13.9 g Wt of oxalic acid in water = 1000 (0.0261) g = 26.1 g That ether layer is removed, and the second jolt of 300 mL fresh ether hits, and Wt of oxalic acid in ether = (300 mL)(1.779x g/mL) = 533.7x g Wt of oxalic acid in water = (1000 ml)(x g/mL) = 1000x g 117
Transcribed Image Text:h extraction I never need ne. See the lly enough. his is so sim- a test tube solve in the ottom, float er, dissolve. to be removed d more fresh ave the same e organic is he theory is our organic emembered. he solubility Chapter 13). kind of a fog ended drop- . Or stir the ents of your on-breaking an react with clogs up the anels. he two liquid not enough nt of one layer ore, you may dissolve this rial does not analysis. You 've heard this acid, and you up, you find some solubilities of oxalic acid as follows: 9.5 g/100 g in water; 23.7 g/100 g in etha- nol; 16.9 g/100 g in diethyl ether. Based on the solubilities, you decide to extract into ethanol from water, forgetting for the moment that ethanol is soluble in water and that you must have two insoluble liquids to carry out an extraction. Chagrined, you forget the ethanol and choose diethyl ether. From the preceding solubility data, we can calculate the distribution, or partition coefficient, for oxalic acid in the water-ether extraction. This coefficient (number) is just the ratio of solubilities of the compound you wish to extract in the two lay- ers. Here, K₁ and = solubility of oxalic acid in ether solubility of oxalic acid in water which amounts to 16.9/9.5, or 1.779. Imagine that you have 40 g of oxalic acid in 1000 mL water and you put that in contact with 1000 mL ether. The oxalic acid distributes itself between the two layers. How much is left in each layer? Well, if we let x g equal the amount that stays in the water, 1.779x g of the acid has to walk over to the ether. And so Wt of oxalic acid in ether= (1000 mL)(1.779x g/mL)=1779x g Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g The total weight of the acid is 40 g (now partitioned between two layers) and SO How to Extract and Wash What 2779x g = 40 g x = 0.0144 Wt of oxalic acid in ether = 1779(0.0144) g = 25.6 g Wt of oxalic acid in water = 1000 (0.0144) g = 14.4 g Now, let's start with the same 40 g of oxalic acid in 1000 mL of water, but this time we will do three extractions with 300 mL of ether. The first 300-mL portion hits, and (300 mL) (1.779x g/mL)=533.7x g Wt of oxalic acid in ether = Wt of oxalic acid in water = (1000 mL)(x g/mL) = 1000x g The total weight of the acid is 40 g (now partitioned between two layers) and 1533.7x g = 40 g x = 0.0261 Wt of oxalic acid in ether = 533.7 (0.0261) g = 13.9 g Wt of oxalic acid in water = 1000 (0.0261) g = 26.1 g That ether layer is removed, and the second jolt of 300 mL fresh ether hits, and Wt of oxalic acid in ether = (300 mL)(1.779x g/mL) = 533.7x g Wt of oxalic acid in water = (1000 ml)(x g/mL) = 1000x g 117
C. The vapor pressure of benzaldehyde at 97 °C is about 60 mm Hg. Assume atmospheric pressure is 760 mm Hg. If you collected 20 g of distillate from your steam distillation, how much of it was benzaldehyde? Show your calculation. ga (MW)APA (MW) BPB gB = Read pages 116-118 in Zubrick. The solubility of benzaldehyde is 1 g/350 mL of water and 1 g/1 mL of ether. How much of the benzaldehyde from 3c was dissolved in the ether layer after your first separation?
Transcribed Image Text:C. The vapor pressure of benzaldehyde at 97 °C is about 60 mm Hg. Assume atmospheric pressure is 760 mm Hg. If you collected 20 g of distillate from your steam distillation, how much of it was benzaldehyde? Show your calculation. ga (MW)APA (MW) BPB gB = Read pages 116-118 in Zubrick. The solubility of benzaldehyde is 1 g/350 mL of water and 1 g/1 mL of ether. How much of the benzaldehyde from 3c was dissolved in the ether layer after your first separation?
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